1 ( _abc ) + 1 + 1 ( _bca ) + 1 + 1 ( _cab ) + 1 is equal to -

Answer: Option A

$$\begin{array}{rl}& \text{Given}\text{Expression}\\ & =\frac{1}{{\log }_{a}bc+{\log }_{a}a}+\frac{1}{{\log }_{b}ca+{\log }_{b}b}+\frac{1}{{\log }_{c}ab+{\log }_{c}c}\\ & =\frac{1}{{\log }_a\left(abc\right)}+\frac{1}{{\log }_b\left(abc\right)}+\frac{1}{{\log }_c\left(abc\right)}\\ & ={\log }_{abc}a+{\log }_{abc}b+{\log }_{abc}c\\ & ={\log }_{abc}\left(abc\right)\\ & =1\end{array}$$